Batch files - number of command line arguments

Question

Just converting some shell scripts into batch files and there is one thing I can t seem to find   and that is a simple count of the number of command line arguments   eg  if you have   myapp foo bar   In Shell       -  2    -  foo bar  0 -  myapp  1 -  foo  2 -  bar   In batch      -  2    lt ---- what command      -  foo bar  0 -  myapp  1 -  foo  2 -  bar   So I ve looked around  and either I m looking in the wrong spot or I m blind  but I can t seem to find a way to get a count of number of command line arguments passed in   Is there a command similar to shell s      for batch files   ps  the closest i ve found is to iterate through the  1s and use  shift   but I need to refernece  1  2 etc later in the script so that s no good

User · Answer

If the number of arguments should be an exact number  less or equal to 9   then this is a simple way to check it   if   2        goto args count wrong if   3        goto args count ok   args count wrong echo I need exactly two command line arguments exit  b 1   args count ok

User · Answer

A robust solution is to delegate counting to a subroutine invoked with call  the subroutine uses goto statements to emulate a loop in which shift is used to consume the  subroutine-only  arguments iteratively   echo off setlocal     Call the argument-counting subroutine with all arguments received     without interfering with the ability to reference the arguments    with  1      later  call  count args        Print the result  echo  ReturnValue  argument s  received      Exit the batch file  exit  b     Subroutine that counts the arguments given     Returns the count in  ReturnValue   count args   set  a ReturnValue   0    count args for      if  1     goto  eof      set  a ReturnValue    1      shift   goto count args for

User · Answer

Googling a bit gives you the following result from wikibooks   set argC 0 for   x in      do Set  A argC  1  echo  argC    Seems like cmd exe has evolved a bit from the old DOS days

User · Answer

Try this   SET  A ARGS COUNT 0     FOR   A in      DO SET  A ARGS COUNT  1     ECHO  ARGS COUNT

User · Answer

The function  getargc below may be what you re looking for    echo off setlocal enableextensions enabledelayedexpansion call  getargc argc    echo Count is  argc  echo Args are    endlocal goto  eof   getargc     set getargc v0  1     set  a   getargc v0    0   getargc l0     if not x 2x  xx           shift         set  a   getargc v0     getargc v0    1          goto  getargc l0           set getargc v0      goto  eof   It basically iterates once over the list  which is local to the function so the shifts won t affect the list back in the main program   counting them until it runs out   It also uses a nifty trick  passing the name of the return variable to be set by the function   The main program just illustrates how to call it and echos the arguments afterwards to ensure that they re untouched   C  Here gt  xx cmd 1 2 3 4 5     Count is 5     Args are 1 2 3 4 5 C  Here gt  xx cmd 1 2 3 4 5 6 7 8 9 10 11     Count is 11     Args are 1 2 3 4 5 6 7 8 9 10 11 C  Here gt  xx cmd 1     Count is 1     Args are 1 C  Here gt  xx cmd     Count is 0     Args are C  Here gt  xx cmd 1 2  3 4 5      Count is 3     Args are 1 2  3 4 5

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Avoids using either shift or a for cycle at the cost of size and readability    echo off setlocal EnableExtensions EnableDelayedExpansion set  a arg idx 1 set  curr arg value    loop1 if  arg idx  GTR 9 goto  done set curr arg label    arg idx  call  get value curr arg value  curr arg label  if defined curr arg value     echo  curr arg label    curr arg value    set  a arg idx  1   goto  loop1    done set  a cnt  arg idx -1 echo argument count   cnt  endlocal goto  eof   get value     set  1  2     Output   count cmdline args bat testing more testing arg3 another arg   1  testing  2  more testing  3  arg3  4  another arg argument count  4   EDIT  The  trick  used here involves    Constructing a string that represents a currently evaluated command-line argument variable  i e    1     2  etc   using a string that contains a percent character      and a counter variable arg idx on each loop iteration  Storing that string into a variable curr arg label  Passing both that string   curr arg label   and a return variable s name  curr arg value  to a primitive subprogram get value  In the subprogram its first argument s   1  value is used on the left side of assignment  set  and its second argument s   2  value on the right  However  when the second subprogram s argument is passed it is resolved into value of the main program s command-line argument by the command interpreter  That is  what is passed is not  for example    4  but whatever value the fourth command-line argument variable holds   another arg  in the sample usage   Then the variable given to the subprogram as return variable  curr arg value  is tested for being undefined  which would happen if currently evaluated command-line argument is absent  Initially this was a comparison of the return variable s value wrapped in square brackets to empty square brackets  which is the only way I know of testing program or subprogram arguments which may contain quotes and was an overlooked leftover from trial-and-error phase  but was since fixed to how it is now

User · Answer

You tend to handle number of arguments with this sort of logic   IF   1      GOTO HAVE 0 IF   2      GOTO HAVE 1 IF   3      GOTO HAVE 2   etc   If you have more than 9 arguments then you are screwed with this approach though   There are various hacks for creating counters which you can find here  but be warned these are not for the faint hearted

User · Answer

The last answer was two years ago now  but I needed a version for more than nine command line arguments  May be another one also does      echo off setlocal  set argc  1 set arg0   0 set argv      LOOP set arg   1 if defined arg      set arg argc     1   set argv   argv    1   set  a argc   1   shift   goto  LOOP     dont count arg0 set  a argc - 1 echo  argc   arg s   for  L   i in  0 1  argc    do     call   SHOW ARG arg  i    arg  i       echo converted to local args call   LIST ARGS  argv   exit  b     LIST ARGS setlocal set argc  0 echo arg0  0    LOOP LIST ARGS set arg   1 if not defined arg  exit  b set  a argc   1 call   SHOW ARG arg argc    1 shift goto  LOOP LIST ARGS     SHOW ARG echo  1  2 exit  b   The solution is the first 19 lines and converts all arguments to variables in a c-like style  All other stuff just probes the result and shows conversion to local args  You can reference arguments by index in any function

[shell] Batch files - number of command line arguments

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